### Calculations in Chemistry: Average Atomic Mass

```Average Atomic Mass

An element can exist in a number of forms,
called isotopes. Isotopes are forms of the
same atom that vary in mass as a result of a
different number of neutrons.
Isotopes of Copper





There are two naturally occurring isotopes of
copper.
One isotope weighs in at 62.93 amu, the
other has a mass of 64.94 amu.
These two isotopes have different
proportions in a natural sample of copper.
The lighter isotope is more common
with 69.09% of all naturally occurring copper
What is the percent abundance of the other
isotope?

The remainder of the atoms, 30.91 %, have a
mass of 64.94 amu.


To find the AVERAGE ATOMIC MASS of an
atom, we take into account all of the isotopes
that exist and the percentage of each type.
The calculation of the average atomic mass
is a WEIGHTED AVERAGE.
Average atomic mass =
Σ (mass of isotope × relative abundance)
Σ = sum or to add them all together



Since there are two isotopes for copper, we
will be adding the contributions of 2
isotopes. (That’s where the Σ sign comes in. )
The relative abundance is simply the
percentage of the isotope, but in decimal
form.
69.09% corresponds to a relative abundance
of 0.6909.
Average atomic mass of copper =
(62.93 amu × 0.6909)
[the mass and abundance of isotope #1]
+
(64.94 amu × 0.3091)
[the mass and abundance of isotope #2]
=
63.55



An AVERAGE atom of copper has a mass of
63.55 amu.
Notice that in this problem, we would predict
that the average is closer to the weight of the
lighter isotope.
This is because the lighter form of copper is
more abundant.
Isotope name
Isotope mass (amu)
Percent abundance
Silver-107
106.90509
51.86
Silver-109
108.90470
remainder
Isotope name
Isotope mass (amu)
Relative abundance
Silicon – 28
27.98
92.21%
Silicon – 29
28.98
4.70%
Silicon – 30
29.97
3.09%
Isotope name
Isotope mass (amu)
percentage
Iron – 54
53.94
5.90%
Iron – 56
55.93
91.72%
Iron – 57
56.94
2.10%
Iron – 58
57.93
0.280%



Silver = 107.87 amu
Silicon = 28.09 amu
Iron = 55.84 amu



Read section 1.4 in the text to support what
we did in class today.
pg 27 #1
Pg 29 #1-5,9
```